# C4 Vectors q.Watch

#1
Hello all, am having a bit of trouble working out this question:

The line l1 has vector equation:

The point A has coordinates (4, 8, a) and point B has coordinates (b, 13, 13). Points A and B lie on l1.

Find the values of a and b.

How on earth do you do this?? Many ta
0
9 years ago
#2
split the vector equation up.
x=8+lambda
y=12+lambda
z=14-lambda

for point A, we know the x coord and y coord. We can use either two equations to work out the value of lambda
y=12+lambda | 8=12 + lambda :. lambda = -4
x=8+lambda | 4=8+lambda :. lambda = -4
now we plug lambda=-4 into the z equation
z=14-lambda = 14-4 = 10. so the z coord at point A is 10. Do the same for point B
0
#3
The only answer given in the mark scheme is:

lamda = -4, therefore a = 18

and

mu = 1, therefore b = 9

but I don't get it. Where does the mu pop up from?
0
9 years ago
#4
They're basically the same. Seeing as we've used lambda=-4 to work out the coords of point A, we need a new variable, mu to work out the coords of point B.
Lambda is used to multiply the direction vector to get to point A.
Mu is used to multiply the direction vector to get to point B.
*Shoddy explanation I know..hope you understand what I'm getting at*
0
#5
(Original post by ogloom)
split the vector equation up.
x=8+lambda
y=12+lambda
z=14-lambda

for point A, we know the x coord and y coord. We can use either two equations to work out the value of lambda
y=12+lambda | 8=12 + lambda :. lambda = -4
x=8+lambda | 4=8+lambda :. lambda = -4
now we plug lambda=-4 into the z equation
z=14-lambda = 14-4 = 10. so the z coord at point A is 10. Do the same for point B
Thank you kind sir!
0
#6
so i got past the obstacle that was part a but now i'm having a bit of trouble with part b...

b) Given that the point O is the origin, and that the point P lies on l1 such that OP is perpendicular to l1, find the coordinates of P.

how would you go about doing this? thanks again for any help!
0
9 years ago
#7
well the position vector of P is simply the line equation. and the direction vector to which the p.v is perp to is lamda(1,1,-1)
0
#8
(Original post by T.P.D-L)
well the position vector of P is simply the line equation. and the direction vector to which the p.v is perp to is lamda(1,1,-1)
sorry, i'm a pretty slow when it comes to vectors(/maths) but how do you know it's the direction vector that's perpendicular? i get so confused as to when you should use the direction vector or the position vector!
0
9 years ago
#9
well it says OP (which is the line equation) is perp to
L1 (which is the direction vector of the line equation). You only take the direction vecotr becuase the first part of the line equation is just a point on L1
0
#10
oh i see! thank you!
0
9 years ago
#11
I thought this was tricky. we know point P lies somewhere on line L1
we know
x=8+lambda
y=12+lambda
z=14 - lambda

at the correct value of lambda = the coords of P.
so vector OP is (8+lambda | 12 + lambda | 14 - lambda).
It is perpendicular to L1, which has direction vector (1|1|-1)
so dot product OP and (1|1|-1) and make it equal to 0. Find the value of lambda which solves this equation, plug it into the vector equation for L1 and you're off!
0
#12
(Original post by ogloom)
I thought this was tricky. we know point P lies somewhere on line L1
we know
x=8+lambda
y=12+lambda
z=14 - lambda

at the correct value of lambda = the coords of P.
so vector OP is (8+lambda | 12 + lambda | 14 - lambda).
It is perpendicular to L1, which has direction vector (1|1|-1)
so dot product OP and (1|1|-1) and make it equal to 0. Find the value of lambda which solves this equation, plug it into the vector equation for L1 and you're off!
ahhh ta very much indeed
0
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